h(x) = x^2 - 4x + 11, this is a parabola that opens up, so it has a minimum value. so the y's or h(x)'s below this minimum value are NOT in the range. Remember what the range is, i said it above.

Let's find the min value. We can find it by completing the square (do you know how to do that)? You're in middle school right? Ok, fine, let's do it a simplier way. The min value occurs at the vertex of the parabola. The x-coordinate of the vertex is given by x = -b/2a where a is the coefficient of x^2, in this case 1, and b is the coefficient of x, in this case -4